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Math Help - Function of a Function Quotient Integration

  1. #1
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    Function of a Function Quotient Integration

    Hi All please can i have some help with this!

    <br /> <br /> <br />
\int {\frac{{e^{\tan ^{ - 1} 3x} }}<br />
{{1 - 9x^2 }}} <br /> <br /> <br />
    Last edited by MexicanGringo; July 29th 2008 at 12:01 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MexicanGringo View Post
    Hi All please can i have some help with this!

    .......arctan3x
    y = e..................
    ..........1 + 9x(squared)

    Apologies for poor formula havn't figured how to do it properly yet
    note that the derivative of arctan 3x is 1/(1 + 9x^2). thus, a substitution of u = arctan 3x will yield int e^u du. i trust you can take it from here?


    EDIT: did i miss something? is LaTeX not working?
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  3. #3
    Eater of Worlds
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    You must have been gone for awhile JH. No, Latex is not working and hasn't been for sometime now.
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    Thanks for the help JHevon, OK not sure if this is correct as the type of problem is the inverse of the usuall f'(x) / f(x)

    y = e^arctan3x
    ......1+9x^2

    u = arctan3x

    du/3 = dx/1+9x^2

    = e^arctan3x/3 + C

    Is this correct?
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  5. #5
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    Hello, MexicanGringo!

    You're answer is correct!


    \int \frac{e^{\arctan3x}}{1+9x^2}\,dx

    We have: . \int e^{\arctan3x}\cdot\frac{dx}{1+9x^2}


    Let: u \:=\:\arctan3x \quad\Rightarrow\quad  du \:=\: \frac{3\,dx}{1 + 9x^2} \quad\Rightarrow\quad \frac{dx}{1+9x^2} \:=\:\frac{1}{3}\,du


    Substitute: . \int e^u \cdot \frac{1}{3}\,du \;=\;\frac{1}{3}\int e^u\,du \;=\;\frac{1}{3}\,e^u + C


    Back-substitute: . \frac{1}{3}\,e^{\arctan3x} + C

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  6. #6
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    Excellent!! Thank you all for your help!!
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